Transactions on Combinatorics ISSN (print): , ISSN (on-line): Vol. 4 No. 2 (2015), pp c 2015 University of Isfahan

Transcription

1 Transactions on Combinatorics ISSN (print): , ISSN (on-line): Vol. 4 No. 2 (2015), pp c 2015 University of Isfahan UNICYCLIC GRAPHS WITH STRONG EQUALITY BETWEEN THE 2-RAINBOW DOMINATION AND INDEPENDENT 2-RAINBOW DOMINATION NUMBERS J. AMJADI, M. CHELLALI, M. FALAHAT AND S. M. SHEIKHOLESLAMI Communicated by Hamid Reza Maimani Abstract. A 2-rainbow dominating function (2RDF) on a graph G = (V, E) is a function f from the vertex set V to the set of all subsets of the set {1, 2} such that for any vertex v V with f(v) = the condition u N(v) f(u) = {1, 2} is fulfilled. A 2RDF f is independent (I2RDF) if no two vertices assigned nonempty sets are adjacent. The weight of a 2RDF f is the value ω(f) = v V f(v). The 2-rainbow domination number γ r2 (G) (respectively, the independent 2-rainbow domination number i r2 (G) ) is the minimum weight of a 2RDF (respectively, I2RDF) on G. We say that γ r2 (G) is strongly equal to i r2(g) and denote by γ r2(g) i r2(g), if every 2RDF on G of minimum weight is an I2RDF. In this paper we characterize all unicyclic graphs G with γ r2 (G) i r2 (G). 1. Introduction Throughout this paper, all graphs considered are finite, undirected, loopless and without multiple edges. We refer the reader to [8, 13] for terminology and notation in graph theory. Specifically, let G be a simple graph with vertex set V = V (G) and edge set E = E(G). For every vertex v V, the open neighborhood N(v) is the set {u V uv E} and the closed neighborhood of v is the set N[v] = N(v) {v}. The degree of a vertex v V is deg G (v) = deg(v) = N(v). A unicyclic graph is a connected graph containing exactly one cycle. A vertex of degree one is called a leaf, and its neighbor is called a support vertex. If v is a support vertex, then L v will denote the set of all leaves adjacent to v. A support vertex v is called a strong support vertex if L v > 1. For r, s 1, a double MSC(2010): Primary: 05C69; Secondary: 05C99. Keywords: 2-rainbow domination number, independent 2-rainbow. domination number, strong equality. Received: 19 May 2014, Accepted: 10 October Corresponding author. 1

2 2 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami star S(r, s) is a tree with exactly two vertices that are not leaves, with one adjacent to r leaves and the other to s leaves. For a positive integer k, a k-rainbow dominating function (krdf) of a graph G is a function f from the vertex set V (G) to the set of all subsets of the set {1, 2,..., k} such that for any vertex v V (G) with f(v) = the condition u N(v) f(u) = {1, 2,..., k} is fulfilled. The weight of a krdf f is the value ω(f) = v V f(v). The k-rainbow domination number of a graph G, denoted by γ rk(g), is the minimum weight of a krdf of G. A γ rk (G)-function is a k-rainbow dominating function of G with weight γ rk (G). Note that 1-rainbow domination number is the classical domination number γ(g). The k-rainbow domination was introduced by Brešar, Henning, and Rall [2] and has been studied by several authors (see for example [3, 6, 7, 12]). A k-rainbow dominating function f is called an independent k-rainbow dominating function (abbreviated IkRDF) on G if the set {v V f(v) } is independent. The independent k-rainbow domination number, denoted by i rk (G), is the minimum weight of an IkRDF on G. An independent k-rainbow dominating function with weight i rk (G) is called an i rk (G)-function. k-rainbow domination number is investigated in [1, 4]. The independent Obviously each independent k-rainbow dominating function is a k-rainbow dominating function, and so γ rk (G) i rk (G). If γ rk (G) = i rk (G), then every i rk (G)-function is also a γ rk (G)-function. However not every γ rk (G)-function is an i rk (G)-function, even when γ rk (G) = i rk (G). We say that γ rk (G) and i rk (G) are strongly equal and denote by γ rk (G) i rk (G), if every γ rk (G)-function is an i rk (G)-function. The strong equality between two parameters was introduced by Haynes and Slater in [11] in the first. Also in [9] and [10], Haynes, Henning and Slater gave constructive characterizations of trees with strong equality between some domination parameters. In this paper, we characterize all unicyclic graphs G with γ r2 (G) i r2 (G). For this aim, we use the constructive characterization of trees T with γ r2 (T ) i r2 (T ) provided recently by Amjadi et al. [1]. Below we present the procedure given in [1] to built such trees. Let F 1 be the family of trees that can be obtained from k 1 disjoint stars K 1,2 by adding either a new vertex v or a path uv and joining the centers of stars to v. Also let F 2 be the family including P 5 and all trees obtained from k 2 disjoint P 3 by adding either a new vertex v or a path uv and joining v to a leaf of each P 3. If T belongs to F 1 F 2 {P 5 } then we call the vertex v, the special vertex of T and if T = P 5, then its support vertices are special vertices of T. We now define recursively a collection F of trees such that K 1,2 F, and if T is any tree in F, then we put in F any tree T that can be obtained from T by any of the following seven operations: Operation O 1 : If z is a strong support vertex of T F, then O 1 adds a new vertex x and an edge xz. Operation O 2 : If z is a vertex of T F, then O 2 adds a new tree T 1 F 1 with special vertex x and an edge xz provided that if x is a support vertex, then γ r2 (T z) γ r2 (T ). Operation O 3 : If z is a strong support vertex of T F, then O 3 adds a path zxy.

3 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami 3 Operation O 4 : If z is a vertex of T F which is adjacent to a support vertex of degree 2, then O 4 adds a path zxy. Operation O 5 : If z is a vertex of T F which is adjacent to a strong support vertex, then O 5 adds a path zxyw. Operation O 6 : If z is a vertex of T F, then O 6 adds new tree T 2 F 2 with special vertex x and an edge xz provided that if x is a support vertex, then γ r2 (T z) γ r2 (T ). Operation O 7 : If z is a vertex of T F such that every γ r2 (T )-function assigns to z, then O 7 adds the double star S(1, 2) and an edge zx where x is a leaf of S(1, 2) whose support vertex has degree 3. Theorem A (Amjadi et al. [1]). Let T be a tree. Then i r2 (T ) γ r2 (T ) if and only if T F {K 1 }. 2. Unicyclic graphs G with γ r2 (G) i r2 (G) We begin by giving some definitions and results that will be useful in our characterization. vertex v V (G) is called an empty vertex if f(v) = for every γ rk (G)-function f. A Observation 2.1. If v V (G) is an empty vertex of G, then γ r2 (G) = γ r2 (G v). Proof. Since v is an empty vertex of G, each γ r2 (G)-function is clearly a 2RDF of G v implying that γ r2 (G v) γ r2 (G). If γ r2 (G v) < γ r2 (G), then let f be a γ r2 (G v)-function and define the function h by h(v) = {1} and h(x) = f(x) for x V (G) {v}. Clearly h is a 2RDF of G of weight γ r2 (G) that leads to a contradiction because v is an empty vertex of G. Thus γ r2 (G) = γ r2 (G v). A pair (x, y) of vertices of a tree T F is called a complementary pair, orc-pair for short, if there exists a γ r2 (T )-function f such that {f(x), f(y)} = {{1}, {2}}. For the pair (x, y), let G Tx,y be the graph obtained from T by adding a new vertex z and edges xz and zy. Further, if z is an empty vertex of G Tx,y, then (x, y) is called a c-pair of type 1, and if there is a γ r2 (G Tx,y )-function that assigns {1, 2} to z, then (x, y) is called a c-pair of type 2. Besides these two cases, (x, y) is called a c-pair of type 3. The following observation is straightforward. Observation 2.2. If T F and x, y V (T ) are a c-pair, then γ r2 (T ) = γ r2 (G Tx,y ). A vertex v of a tree T is said to be an independent vertex if (i) every γ r2 (T )-function f with f(v) = 1 is independent and there is at least one γ r2 (T )-function with this property, (ii) there is no γ r2 (T )-function f such that f(v) = 2. We denote by H the set of all trees in F having an independent vertex. In the next, we provide a constructive characterization of unicyclic graphs G with γ r2 (G) i r2 (G). Let U be the family of graphs G such that G is obtained from some trees in F by one of the operations T 1, T 2, T 3, or T 4 or obtained from even number of trees in H by operation T 5. Operation T 1 : Let x and y be two non-adjacent vertices of T F. The Operation T 1 adds the edge xy if the following two conditions hold:

4 4 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami (1) every γ r2 (T )-function assigns to at least one of the x and y, and (2) T u, (u {x, y}) has no 2RDF f of weight γ r2 (T ) such that f is not independent and v (N T (u) ({x,y} {u})) f(v) = {1, 2}. Operation T 2 : Assume T 0, T 1,..., T t F, x, y V (T 0 ) and v i V (T i ) for i {1,...,t} and let (1) (x,y) is a c-pair of type 1, (2) for each i {1,...,t}, v i is an empty vertex of T i. Then Operation T 2 adds a new vertex z and edges zx, zy and zv i for each i {1,...,t}. Operation T 3 : Assume T 0, T 1,..., T t F, x, y V (T 0 ) and v i V (T i ) for i {1,...,t} and let (1) (x,y) is a c-pair of type 2, (2) T 0 {x, y} has no 2RDF f of weight γ r2 (T 0 ) 1 that is not independent and j ( u NT0 (x) {y}f(u)) ( u NT0 (y) {x}f(u)) for some j {1, 2}, (3) for each i {1,...,t}, v i is an empty vertex of T i, (4) T i v i F for each i {1,...,t}. Then Operation T 3 adds a new vertex z and edges zx, zy and zv i for i {1,...,t}. Operation T 4 : Assume T 0, T 1,..., T t F, x, y V (T 0 ) and v i V (T i ) for i {1,...,t} such that (1) (x,y) is a c-pair of type 3, (2) T 0 {x, y} has no 2RDF f of weight γ r2 (T 0 ) 1 that is not independent and j ( u NT0 (x) {y}f(u)) ( u NT0 (y) {x}f(u)) for j {1, 2}, (3) v i is an empty vertex of T i for each i {1,...,t}, (4) for each i {1,...,t}, T i v i has no 2RDF g of weight γ r2 (T i ) such that g is not independent and 1 u NTi (v i )g(u) or 2 u NTi (v i )g(u). Then Operation T 4 adds a new vertex z and edges zx, zy and zv i for i {1,...,t}. Operation T 5 : Let t 0 be an even integer, T 1, T 2,..., T t H, and v i V (T i ) be an independent vertex for each i {1,..., t}. Then Operation T 5 adds new vertices u 1, u 2,..., u t and edges u i v i, for each i {1,..., t}, v i u i+1 for each i {1,..., t 1} and the edge v t u 1. Lemma 2.3. If T is a tree with γ r2 (T ) i r2 (T ) and G is a unicyclic graph obtained from T by Operation T 1, then γ r2 (G) i r2 (G). Proof. Let T F and let x and y be non-adjacent vertices of T jointed by Operation T 1. Clearly every γ r2 (T )-function is a 2RDF of G and hence γ r2 (G) i r2 (G) γ r2 (T ) = i r2 (T ). Now we show that γ r2 (G) = γ r2 (T ). Suppose to the contrary that γ r2 (G) < γ r2 (T ) and let f be a γ r2 (G)-function. If f(x) = f(y) =, or {f(x), f(y)}, then f is a 2RDF of T with weight less than γ r2 (T ), a contradiction. If f(x) and f(y) = (the case f(x) = and f(y) is similar), then the function g defined by g(y) = {1}, g(w) = f(w) for w V (G) {y} is a γ r2 (T )-function which contradicts (i). Hence γ r2 (G) = i r2 (G) = γ r2 (T ). It will now be shown that γ r2 (G) i r2 (G). Suppose g is a γ r2 (G)-function that is not independent. If g(x) and g(y) then g is a γ r2 (T )-function, that contradicts (i). If g(x) = g(y) =, then g is a γ r2 (T )-function that is not independent, a contradiction with T F. If g(x) and g(y) = (the case g(x) = and g(y) is similar), then g is a 2DRF of

5 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami 5 T y of weight γ r2 (T ) that is not independent and v (N(y) {x}) g(v) = {1, 2}, a contradiction with (ii). This completes the proof. Lemma 2.4. Let T 0, T 1,..., T t F, x, y V (T 0 ) be a c-pair of type 1 and v i V (T i ) is an empty vertex of T i for every i {1,..., t}. If G is a unicyclic graph obtained from T 0, T 1,..., T t by Operation T 2, then γ r2 (G) i r2 (G). Proof. Assume z is the new vertex added by Operation T 2 for obtaining G. First we show that γ r2 (G) = i r2 (G). Let f i be a γ r2 (T i )-function for every i {0,..., t} and let {f 0 (x), f 0 (y)} = {{1}, {2}}. Define h : V (G) P({1, 2}) by h(z) = and h(u) = f i (u) for u V (T i ) and 0 i t. Since T i F, we deduce that h is an I2RDF of G with weight t i=0 γ r2(t i ) implying that (2.1) γ r2 (G) i r2 (G) γ r2 (T i ) = i r2 (T i ). i=0 Assume g is a γ r2 (G)-function. We claim that g(z) =. Suppose to the contrary that g(z). If ω(g V (Tj )) < γ r2 (T j ) for some 1 j t, then the function f defined by f(v j ) = {1} and f(u) = g(u) for u V (T j ) {v j } is a 2RDF of T j with weight at most γ r2 (T i ) which leads to a contradiction because v j is an empty vertex of T j. Thus ω(g V (Tj )) γ r2 (T j ) for each j {1,..., t}. It follows from (2.1) that ω(g V (T0 ) {z}) γ r2 (T 0 ). This implies that g V (GT0 ) is a γ r2 (G T0 )-function with xy xy g(z), a contradiction. Hence g(z) =. Thus g V (Ti ) is a 2RDF for each i {0,..., t}. Therefore ω(g V (Ti )) γ r2 (T i ) = i r2 (T i ) for every i {0,..., t}. Hence γ r2 (G) = ω(g) t i=0 γ r2(t i ) and by (2.1) we have γ r2 (G) = i r2 (G) = t i=0 γ r2(t i ). Now it will be shown that γ r2 (G) i r2 (G). Assume h is a γ r2 (G)-function that is not independent. Using an argument similar to that described above, shows that h(z) = and the function h V (Ti ) is a γ r2 (T i )-function for each i. Since h is not independent, we deduce that h V (Ti ) is not independent for some i, a contradiction with T i F. This completes the proof. Lemma 2.5. Let T 0, T 1,..., T t F, x, y V (T 0 ) and v i V (T i ) for 1 i t satisfy in the condition of Operation T 3. If G is a unicyclic graph obtained from T 0, T 1,..., T t by Operation T 3, then γ r2 (G) i r2 (G). Proof. Suppose z is the vertex added by Operation T 3 for obtaining G. We first show that γ r2 (G) = i r2 (G) = t i=0 γ r2(t i ). As in the proof of Lemma 2.4, we can see that (2.2) γ r2 (G) i r2 (G) γ r2 (T i ) = i r2 (T i ). i=0 Assume g is a γ r2 (G)-function. As in the proof of Lemma 2.4, we have ω(g V (Tj )) γ r2 (T j ) for each j {1,..., t}. On the other hand, the function g, restricted to G T0x,y is a 2RDF of G T0x,y implying that ω(g ) γ GT0 r2(t 0x,y ) = γ r2 (T 0 ) by Observation 2.2. Thus γ r2 (G) t x,y i=0 γ r2(t i ). It follows from (2.2) that γ r2 (G) = i r2 (G) = t i=0 γ r2(t i ) = t i=0 i r2(t i ). Now we show that γ r2 (G) i r2 (G). Assume h is a γ r2 (G)-function that is not independent. By above argument, h V (Ti ) is a γ r2 (T i )- function for each i {1,..., t}. We consider two cases. i=0 i=0

6 6 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami Case 1. h(z) =. Since h is not independent, h V (Ti ) for some i, is a γ r2 (T i )-function that is not independent, a contradiction with T i F. Case 2. h(z). If h(v i ) for some 1 i t, then h V (Ti ) is a γ r2 (T i )-function and v i is not an empty vertex of T i, a contradiction. Thus h(v i ) = for every i {1,..., t}. This implies that h V (Ti ) {v i } is a γ r2 (T i v i )-function, by Observation 1. If h V (Ti ) {v i } is not independent for some i, we obtain a contradiction with T i v i F. Henceforth, we let h V (Ti ) {v i } is independent for each i. Thus h GT0 x,y is a γ r2 -function that is not independent. Consider two subcases. Subcase 2.1. h(z) = {1, 2}. If h(x) (the case h(y) is similar), then the function g defined by g(y) = {1}, g(u) = h(u) for u V (T 0 ) {x} is a 2RDF of T 0 of weight less than γ r2 (T 0 ), a contradiction. Therefore, we assume h(x) = h(y) =. Then the function g defined by g(x) = g(y) = {1}, g(u) = h(u) for u V (T 0 ) {x, y} is a γ r2 (T 0 )-function that is not independent, a contradiction. Subcase 2.2. h(z) = 1. We may assume without loss of generality that h(z) = {1}. If {h(x), h(y)}, then h V (T0 ) is a 2RDF of weight less than γ r2 (T 0 ), a contradiction. If h(x) and h(y) = (the case h(x) = and h(y) is similar), then 2 u NT0 (y)h(u) and the function g defined by g(y) = {1} and g(u) = h(u) for u V (T 0 ) {y} is a 2RDF of weight γ r2 (T 0 ) that is not independent, a contradiction with T 0 F. Hence, let h(x) = h(y) =. Then 2 u N T0 (v) h(u) for v {x, y}. If u N T0 (x)h(u) = {1, 2} (the case u NT0 (y)h(u) = {1, 2} is similar), then the function g defined by g(y) = {1} and g(u) = h(u) for u V (T 0 ) {y} is a 2RDF of weight γ r2 (T 0 ) that is not independent which is a contradiction again. Thus u NT0 (x)h(u) = u NT0 (y)h(u) = {2}. But then h V (T0 ) {x,y} is a 2RDF on T 0 {x, y} of weight γ r2 (T 0 ) 1 that is not independent contradicting the assumption. This completes the proof. Lemma 2.6. Let T 0, T 1,..., T t F, x, y V (T 0 ) and v i V (T i ) for each i {1,..., t} satisfy in the conditions of Operation 4. If G is the unicyclic graph obtain from T 0, T 1,..., T t by Operation T 4, then γ r2 (G) i r2 (G). Proof. Suppose z is the vertex added by Operation T 4 to obtain G. As in the proof of Lemma 2.4, we obtain (2.3) γ r2 (G) i r2 (G) γ r2 (T i ). Assume g is a γ r2 (G)-function. If g(z) = then g V (Ti ) is a 2RDF of T i for every i {0,..., t} and hence ω(g V (Ti )) γ r2 (T i ) for each i {0,..., t}. Therefore γ r2 (G) = ω(g) t i=0 γ r2(t i ). Assume g(z). Then g V (T0 ) {z} is a 2RDF of G T0 xy and for each i either g V (T i ) is a 2RDF of T i or g V (Ti ) {v i } is a 2RDF of T i v i. By Observations 2.1 and 2.2, we deduce that γ r2 (G) t i=0 γ r2(t i ). It follows from (2.3) that γ r2 (G) = i r2 (G) = t i=0 γ r2(t i ). Now we show that γ r2 (G) i r2 (G). Suppose to the contrary that g is a γ r2 (G)-function that is not independent. If g(z) =, then as in the proof of Case i=0

7 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami 7 1 in Lemma 2.5 we obtain a contradiction. Let g(z). Since x and y are a c-pair of type 3, we have g(z) = 1. Assume, without loss of generality, that g(z) = {1}. Since g is not independent, there are two vertices u, v V (G) such that {f(u), f(v)}. Since ω(g V (Ti )) = γ r2 (T i ) and v i is an empty vertex of T i, we deduce that g(v i ) = for each 1 i t. It follows that 2 w NTi (v i )g(w) for each 1 i t, and u, v V (G T0x,y ) or u, v V (T i v i ) for some 1 i t. By condition 3., we deduce that u, v V (G ). If g(x) = g(y) =, then we must have 2 ( T0 x,y w N T0 (x)f(w)) ( w NT0 (y)f(w)) and u, v V (T 0 )\{x, y} contradicting condition 2. Let without loss of generality that f(x). Then the function h : V (G) P({1, 2}) defined by h(y) = {1} and h(w) = g(w) for w V (T 0 ) {y}, is a γ r2 (T 0 )-function that is not independent, a contradiction. Thus γ r2 (G) i r2 (G) and the proof is complete. Lemma 2.7. Assume t is an even integer, T 1, T 2,..., T t H, and v i V (T i ) is an independent vertex for each i {1,..., t}. If G is the unicyclic graph obtained from T 1,..., T t by Operation T 5, then γ r2 (G) i r2 (G). Proof. We use the same notation as defined in Operation T 5. Suppose f i is an independent γ r2 (T i )- function such that f(v i ) = 1 for each i. We may assume that f(v i ) = {1} if i is odd and f(v i ) = {2} if i is even. Define the function f : V (G) P({1, 2}) by f(u) = f i (u) for u V (T i ) and f(u i ) = for each i {1,..., t}. Clearly f is an I2RDF of G of weight t i=1 γ r2(t i ) implying that (2.4) γ r2 (G) i r2 (G) γ r2 (T i ). i=1 It will now be shown that γ r2 (G) = t i=1 γ r2(t i ). Assume to the contrary that γ r2 (G) < t i=1 γ r2(t i ) and let f be a γ r2 (G)-function such that t i=1 f(u i) is as small as possible. We claim that f(u i ) {1, 2} for each i. Suppose to the contrary that f(u i ) = {1, 2} for some i, say i = 2. Then the function g : V (G) P({1, 2}) defined by g(u 2 ) =, g(v 1 ) = f(v 1 ) {1}, g(v 2 ) = f(v 2 ) {2} and g(x) = f(x) otherwise, is a 2RDF of G such that t i=1 g(u i) < t i=1 f(u i), a contradiction. Hence f(u i ) {1, 2} for each i. Since γ r2 (G) < t i=1 γ r2(t i ), we deduce that ω(f V (Ti ) {u i+1 }) < γ r2 (T i ) for some 1 i t, say i = 1. If f V (T1 ) is a 2RDF of T 1, then its weight is less than γ r2 (T 1 ) which is a contradiction. Let f V (T1 ) is not a 2RDF of T 1. This implies that f(v 1 ) =. If f(u 2 ), then f(u 2 ) = 1 and the function h : V (T 1 ) P({1, 2}) defined by h(v 1 ) = f(u 2 ), h(x) = f(x) for x V (T 1 ) {v 1 } is a 2RDF of T 1 of weight less than γ r2 (T 1 ) which is a contradiction again. Thus f(u 2 ) =. Now to rainbowly dominate v 1, we must have f(u 1 ). Since f(u 1 ) {1, 2}, we may assume without loss of generality that f(u 1 ) = {1}. It follows that 2 u NT1 (v 1 )f(u). But then the function h 1 : V (T 1 ) P({1, 2}) defined by h 1 (v 1 ) = {1}, h 1 (x) = f(x) for x V (T 1 ) {v 1 } is a γ r2 (T 1 )-function such that h 1 is not independent and h 1 (v 1 ) = 1, a contradiction by T 1 H. Thus ω(f V (Ti ) {u i+1 }) γ r2 (T i ) for each i {1,..., t} and hence γ r2 (G) t i=1 γ r2(t i ). It follows from (2.4) that γ r2 (G) = t i=1 γ r2(t i ) implying that γ r2 (G) = i r2 (G). Now we show that γ r2 (G) i r2 (G). Suppose to the contrary that g is a γ r2 (G)-function that is not independent and let u and v be two vertices such that uv E(G) and {g(u), g(v)}. We consider two cases.

8 8 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami Case 1. u, v V (T i ) for some 1 i t. If g(v i ) = 1 then g V (Ti ) is a γ r2 (T i )-function that is not independent, contradicting (i) of the definition of independent vertex. If g(v i ) = 2, then obviously g V (Ti ) is a 2RDF of T i. Since T i H, it follows from the property (2) that g V (Ti ) is not a γ r2 (T i )-function which implies ω(g V (Ti ) {u i+1 }) ω(g V (Ti )) > γ r2 (T i ). Since ω(g V (Ti ) {u i+1 }) γ r2 (T i ) for each 1 i t, we obtain γ r2 (G) > t i=1 γ r2(t i ), a contradiction. Henceforth, we assume g(v i ) = 0. If ω(g V (Ti )) < γ r2 (T i ) then we define h(v i ) = {1} and h(w) = g(w) for w V (T i ) {v i } to obtain a γ r2 (T i )-function that is not independent, a contradiction with the property (1). Let ω(g V (Ti )) = γ r2 (T i ). Then we must have g(u i+1 ) =, otherwise ω(g V (Ti ) {v i+1 }) ω(g V (Ti )) > γ r2 (T i ) and we obtain a contradiction as above. It follows from g(u i+1 ) = and g(v i ) = that g(v i+1 ) = {1, 2}. By the property (2), we have ω(g V (Ti+1 ) {u i+2 }) > γ r2 (T i+1 ) which leads to a contradiction as above. Case 2. {v, u} = {v i, u i+1 } for some i. (the case {v, u} = {v i, u i } is similar). Assume, without loss of generality, that v = v i and u = u i+1. Since g V (Ti ) is a γ r2 (T i )-function and since g(u i+1 ), we deduce that ω(g V (Ti ) {u i }) > γ r2 (T i ). Since ω(g V (Tj ) {u j+1 }) γ r2 (T j ) for each 1 j t, we obtain γ r2 (G) > t i=1 γ r2(t i ), a contradiction. Thus g is independent and hence γ r2 (G) i r2 (G). This completes the proof. In the next, we show that if G is a unicyclic graph with i r2 (G) γ r2 (G), then G U. Theorem 2.8. If G is a unicyclic graph of order n 3, with i r2 (G) γ r2 (G), then G U. Proof. Let C = (u 1, u 2,..., u k ) be the unique cycle of G and let C 0 = {u V (C) there is a γ r2 (G) function that assigns to u}. Clearly C 0 because γ r2 (G) i r2 (G). Also let C 1 = {u i C 0 there is a γ r2 (G) function f with f(u i ) = such that f(u i 1 ) f(u i+1 ) or f(u i+1 ) f(u i 1 )}. We consider two cases. Case 1. C 1. Assume, without loss of generality, that u 2 C 1. By assumption, G has an i r2 (G)-function f such that f(u 2 ) = and f(u 1 ) f(u 3 ) or f(u 3 ) f(u 1 ). Suppose, without loss of generality, that f(u 1 ) f(u 3 ). Let T = G u 1 u 2. Clearly T is a tree. We show that T F and G can be obtained from T by Operation T 1. Obviously f is an I2RDF of T that yields γ r2 (T ) i r2 (T ) γ r2 (G). On the other hand, every γ r2 (T )-function is a 2RDF of G implying that γ r2 (T ) γ r2 (G). Therefore γ r2 (T ) = i r2 (T ) = γ r2 (G). If T has a non independent γ r2 (T )-function f, then f is γ r2 (G)-function that is not independent, a contradiction with i r2 (G) γ r2 (G). This yields i r2 (T ) γ r2 (T ). If T has a γ r2 (T )-function g such that {g(u 1 ), g(u 2 )}, then g is a γ r2 (G)-function that is not independent, a contradiction again. Hence T satisfies (i) of Operation T 1. Suppose u {u 1, u 2 }. If T u has a 2RDF g of weight γ r2 (T ) such that g is not independent and v NT (u) ({u 1,u 2 } {u})g(v) = {1, 2}, then the function h : V (G) P({1, 2}) defined by h(u) = and h(w) = g(w) for w V (G) {u} is a 2RDF of G of weight γ r2 (G) that is not independent, a contradiction with i r2 (G) γ r2 (G). Therefore T satisfies (ii) of Operation T 1. Thus G can be obtained from T by applying Operation T 1.

9 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami 9 Case 2. C 1 =. Assume, without loss of generality, that u 2 C 0 and deg(u 2 ) deg(x) for every x C 0. Since C 1 =, for every γ r2 (G)-function g with g(u 2 ) =, we have g(u 1 ) g(u 3 ) and g(u 3 ) g(u 1 ). This implies {g(u 1 ), g(u 3 )} = {{1}, {2}}. Let f be a γ r2 (G)-function with f(u 2 ) =, f(u 1 ) = {1} and f(u 3 ) = {2}. We consider the following subcases. Subcase 2.1. deg(u 2 ) 3. Let N(u 2 ) {u 1, u 3 } = {v 1, v 2,..., v t } and let T 0, T 1,..., T t be the component of G u 2 such that u 1, u 3 V (T 0 ) and v i V (T i ) for each i {1,..., t}. If f(v j ) for some 1 j t, then f(u 1 ) f(v j ) or f(u 3 ) f(v j ), say f(u 1 ) f(v j ), and an argument similar to that described in Case 1, shows that G can be obtained from T = G u 1 u 2 by Operation T 1. Henceforth, we assume that f(v i ) = for each i {1,..., t}. Therefore we may assume for any γ r2 (G)-function g with g(u 2 ) =, g(v i ) = for each i {1,..., t}. We claim that T i F, ω(f V (Ti )) = γ r2 (T i ) and v i is an empty vertex of T i for each i {1,..., t}. Since i r2 (G) γ r2 (G) and since f(u 2 ) =, the function f V (Ti ) is an I2RDF of T i and hence γ r2 (T i ) i r2 (T i ) ω(f V (Ti )) for each 0 i t. If γ r2 (T i ) < ω(f V (Ti )) for some i, then let g 1 be a γ r2 (T i )-function and define h : V (G) P({1, 2}) by h(u) = f(u) for u V (G) V (T i ) and h(u) = g 1 (u) for u V (T i ). Obviously h is a 2RDF of G with weight less than γ r2 (G), a contradiction. So γ r2 (T i ) = ω(f V (Ti )) for each i {0,..., t}. If T i F for some i, then let g 2 be a γ r2 (T i )-function that is not independent and define h 1 : V (G) P({1, 2}) by h 1 (u) = f(u) for u V (G) V (T i ) and h 1 (u) = g 2 (u) for u V (T i ). Clearly h 1 is a γ r2 (G)-function that is not independent, a contradiction with γ r2 (G) i r2 (G). Similarly, we can see that v i is an empty vertex of T i for each i {1,..., t}. If u 2 is an empty vertex of G then u 1 and u 3 are a c-pair of type 1 of T 0 and G can be obtained by using Operation T 2. Now suppose there is a γ r2 (G)-function h that assigns {1, 2} to u 2. Then u 1 and u 3 are a c-pair of type 2. If T i v i, for some 1 i t, has a 2RDF g of weight γ r2 (T i ) that is not independent, then define h 1 (u) = h(u) for u V (G) V (T i ) and h 1 (u) = g(u) for u V (T i v i ) and h 1 (v i ) = to obtain a non independent γ r2 (G)-function, a contradiction with γ r2 (G) i r2 (G). Thus T i v i F for each i {1,..., t}. If T 0 {u 1, u 3 } has a 2RDF g of weight γ r2 (T 0 ) 1 such that g is not independent and j ( u NT0 (u 1 )g(u)) ( u NT0 (u 3 )g(u)) for some j {1, 2}, then define h 2 (u) = g(u) for u V (T 0 ) {u 1, u 3 }, h 2 (u 1 ) = h 2 (u 3 ) =, h 2 (u 2 ) = {1, 2} {j} and h 2 (u) = g i (u) for u V (T i ), 1 i t, where g i is arbitrary γ r2 (T i )-function to obtain a non independent γ r2 (G)- function, a contradiction again. Thus the conditions of Operation T 3 holds and G can be obtain from T 0, T 1,..., T t by Operation T 3. Finally let, there is a γ r2 (G)-function h that assign {1} (the case {2} is similar) to u 2. Then u 1 and u 3 are a c-pair of type 3. Since γ r2 (G) i r2 (G), h(v i ) = for each 1 i t. It follows that 2 N Ti (v i ) for each i {1,..., t}. If ω(h V (Ti )) < γ r2 (T i ) for some i, then the function k : V (T i ) P({1, 2}) defined by k(v i ) = {1} and k(w) = h(w) for w V (T i ) {v 1 }, is a γ r2 (T i )-function that is not independent, a contradiction. Hence, ω(h V (Ti )) = γ r2 (T i ) for each i. We claim that for each i, T i v i has no 2RDF g of weight γ r2 (T i ) that it is not independent and j u NTi (v i )g(u) for some j {1, 2}. Assume to the contrary that T i v i has an 2RDF g of weight γ r2 (T i ) such that j u NTi (v i )g(u) (j {1, 2}) and g is not independent for some i. If j = 2, then the

10 10 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami function h 3 defined by h 3 (v i ) =, h 3 (u) = h(u) for u V (G) V (T i ) and h 3 (u) = g(u) for u V (T i ), is a γ r2 (G)-function that is not independent, a contradiction with γ r2 (G) i r2 (G). If j = 1, the defined h 4 by h 4 (v i ) =, h 4 (u) = h(u) for u V (G) V (T i ), h 4 (u) = {1, 2} \ g(u) if u V (T i ) and g(u) = 1, and h 4 (u) = g(u) when u V (T i ) and g(v i ) 1, to obtain a γ r2 (G)-function that is not independent, a contradiction again. This proves the claim. As above we can see that T 0 {u 1, u 3 } has no 2RDF g of weight γ r2 (T 0 ) 1 that it is not independent and j ( u NT0 (u 1 )f(u)) ( u NT0 (u 2 )f(u)) for some j {1, 2}. Therefore G can be obtained from T 0, T 1,..., T t by Operation T 4. Subcase 2.2. deg(u 2 ) = 2. Assume, without loss of generality, that f(u 1 ) = {1} and f(u 3 ) = {2}. Since γ r2 (G) i r2 (G), f is independent and hence f(u 4 ) =. By the choice of u 2 and the fact that C 1 =, we have deg(u 4 ) = 2 and f(u 5 ) = {1}. Repeating this process, we deduce that f(u 2m ) =, f(u 2m 1 ) = {1} and f(u 2m+1 ) = {2} for each integer m 1. It follows that k = 4s for some s and and every vertex of C with even index belongs to C 0. By the choice of u 2, we deduce that deg(u 2 ) = = deg(u k ) = 2. First let C 0 {u 2, u 4,..., u k } =. Assume, without loss of generality, that u 3 C 0 {u 2,..., u k }. By the choice of u 2, deg(u 3 ) = 2. Using an argument similar to that described in this subcase, we obtain {u 1, u 3,..., u k 1 } C 0 and deg(u 1 ) = = deg(u k 1 ) = 2. Since G is connected, we deduce that G = C 4s. Now we can obtain G from 2s single vertices by Operation T 5. Now let C 0 = {u 2, u 4,..., u k }. Assume T 1, T 3,..., T 4s 1 are the components of G {u 2,..., u k } and u 2m 1 V (T 2m 1 ) for 1 m 2s. First we show that γ r2 (T 2m 1 ) = ω(f V (T2m 1 )) for each m. Clearly f V (T2m 1 ) is an independent 2RDF of T 2m 1 and hence γ r2 (T 2m 1 ) ω(f V (T2m 1 )) for each m. Assume to the contrary that γ r2 (T 2m 1 ) < ω(f V (T2m 1 )) for some m and let g be a γ r2 (T 2m 1 )- function. If g(v 2m 1 ) 1, then we can assume that f(v 2m 1 ) g(v 2m 1 ). Define the function h by h(u) = g(u) if u V (T 2m 1 ) and h(u) = f(u) otherwise, to obtain a 2RDF of G of weight less than γ r2 (G), a contradiction. If g(v 2m 1 ) = 0, then u NT2m 1 (v 2m 1 ) g(u) = 2. Define the function h by h(u) = f(u) for u V (G) V (T 2m 1 v 2m 1 ) and h(u) = g(u) otherwise to obtain a γ r2 (G)-function that is not independent, a contradiction with γ r2 (G) i r2 (G). Thus γ r2 (T 2m 1 ) = ω(f V (T2m 1 )) for each i. Now we show that T 1, T 3,..., T 4s 1 H. Note that f V (T2m 1 ) is an i r2 (T 2m 1 )-function with f(v 2m 1 ) = 1 for 1 m 2s. Let g be an i r2 (T 2m 1 )-function such that g(v 2m 1 ) = 1. Let without loss of generality that g(v 2m 1 ) = f(v 2m 1 ). Then the function h defined by h(u) = g(u) if u V (T 2m 1 ) and h(u) = f(u) otherwise, is an i r2 (G)-function. Since γ r2 (G) i r2 (G), h is independent and hence g is independent. Thus T 2m 1 satisfies (i) in the definition of independent vertex. If there is a γ r2 (T 2m 1 )-function g with g(v 2m 1 ) = 2, then v 2m 2, v 2m C 1 a contradiction with C 1 =. Thus T 2m 1 satisfies (ii) in the definition of independent vertex, for each m. Thus v 2m 1 is an independent vertex of T 2m 1 and hence T 1, T 3,..., T 4s 1 H. Thus G can be obtain from T 1, T 3,..., T 4s 1 by Operation T 5. This completes the proof.

11 Trans. Comb. 4 no. 2 (2015) 1-11 J. Amjadi, M. Chellali, M. Falahat and S. M. Sheikholeslami 11 Now we are ready to state the main theorem of this paper. Theorem 2.9. Let G be a unicyclic graph. Then i r2 (G) γ r2 (G) if and only if G U. References [1] J. Amjadi, M. Falahat, N. Jafari Rad and S. M. Sheikholeslami, Strong equality between the 2-rainbow domination and independent 2-rainbow domination numbers in trees, Bull. Malays. Math. Sci. Soc., (to appear). [2] B. Brešar, M. A. Henning and D. F. Rall, Rainbow domination in graphs, Taiwanese J. Math., 12 (2008) [3] B. Brešar and T. K. Šumenjak, On the 2-rainbow domination in graphs, Discrete Appl. Math., 155 (2007) [4] M. Chellali and N. Jafari Rad, Independent 2-rainbow domination in graphs, J. Combin. Math. Combin. Comput., (to appear). [5] M. Chellali and N. Jafari Rad, Strong equality between the Roman domination and independent Roman domination numbers in trees, Discuss. Math. Graph Theory, 33 (2013) [6] N. Dehgardi, S. M. Sheikholeslami and L. Volkmann, The rainbow domination subdivision numbers of graphs, Mat. Vesnik, (to appear). [7] M. Falahat, S. M. Sheikholeslami and L. Volkmann, New bounds on the rainbow domination subdivision number, Filomat, 28 (2014) [8] T. W. Haynes, S. T. Hedetniemi and P. J. Slater, Fundamentals of Domination in graphs, Marcel Dekker, Inc., New York, [9] T. W. Haynes, M. A. Henning and P. J. Slater, Strong equality of domination parameters in trees, Discrete Math., 260 (2003) [10] T. W. Haynes, M. A. Henning and P. J. Slater, Strong equality of upper domination and independence in trees, Util. Math., 59 (2001) [11] T. W. Haynes and P. J. Slater, Paired-domination in graphs, Networks, 32 (1998) [12] S. M. Sheikholeslami and L. Volkmann, The k-rainbow domatic number of a graph, Discuss. Math. Graph Theory, 32 (2012) [13] D. B. West, Introduction to Graph Theory, Prentice-Hall, Inc., J. Amjadi Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran j-amjadi@azaruniv.edu M. Chellali LAMDA-RO Laboratory, Department of Mathematics, University of Blida, B. P. 270, Blida, Algeria m chellali@yahoo.com M. Falahat Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran mofalahat@yahoo.com S. M. Sheikholeslami Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran s.m.sheikholeslami@azaruniv.edu